Detection limit. In spectrophotometry, we measure the concentration of an analyte by its absorbance of light. A low-concentration sample was prepared and nine replicate measurements gave absorbances of 0.0047,0.0054,0.0062,0.0060,0.0046,0.0056,0.0052,0.0044, and 0.0058. Nine reagent blanks gave values of 0.0006,0.0012, 0.0022,0.0005,0.0016,0.0008,0.0017,0.0010, and 0.0011.

a) Find the absorbance detection limit with equation 5-3.

b) The calibration curve is a graph of absorbance versus concentration. Absorbance is a dimensionless quantity. The slope of the calibration curve is $\mathbf{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{24}\mathbf{}\mathbf{x}\mathbf{}{\mathbf{10}}^{\mathbf{4}}\mathbf{}{\mathbf{M}}^{\mathbf{-}\mathbf{1}}$Find the concentration detection limit with Equation 5-5.

(c) Find the lower limit of quantitation with Equation 5-6.

Detection limit:

This is the concentration of an analyte that produces a signal which is equal to three times the standard deviation of a signal from a blank.

The absorbance detectable concentration ${\mathrm{y}}_{\mathrm{dl}}$can be defined as,:

signal detection limit${\mathrm{y}}_{\mathrm{dl}}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s}$

Minimum detectable concentration$=\frac{3\mathrm{s}}{\mathrm{m}}$

Lower limit of quantitation$=\frac{10\mathrm{s}}{\mathrm{m}}$

Where,

$\mathrm{s}=$Standard deviation.

$\mathrm{m}=$Slope of linear calibration curve.

a)The low concentrations samples near the detection limit are.

0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011.

The mean reading of the nine blanks is 0.001189.

The calculated standard deviation of the nine samples is 0.000644.

$$\begin{gathered} {\mathrm{y}}_{\mathrm{dl}}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ {\mathrm{y}}_{\mathrm{dl}}=0.00118+3\left(0.00688\right) \\ =0.022 \end{gathered}$$

b)The low concentrations samples near the detection limit are. 0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011.

The mean reading of nine blanks is 0.001189.

The slope of the calibration curve is $2.24\mathrm{x}{10}^4{\mathrm{M}}^1$

The calculated standard deviation of the nine samples is 0.000644.$\mathrm{minimum}\mathrm{detectable}\mathrm{concentration}=\frac{3\mathrm{s}}{\mathrm{m}}$

$$\begin{gathered} =\frac{3\left(0.000644\right)}{2.24\mathrm{x}{10}^4{\mathrm{M}}^{-1}} \\ =8.6\mathrm{x}{10}^{-8}\mathrm{M} \end{gathered}$$

c)The low concentrations samples near the detection limit are.

0.0006,0.0012,0.0022,0.0005,0.0016,0.0008,0.00017,0.0010,0.0011.

The mean reading of nine blanks is 0.001189.

The slope of the calibration curve is $2.24\mathrm{x}{10}^4{\mathrm{M}}^{-1}$.

The calculated standard deviation of the nine samples is 0.000644.

lower limit of quantitation

$$\begin{gathered} \mathrm{Lower}\mathrm{limit}\mathrm{of}\mathrm{quantitation}=\frac{10\mathrm{s}}{\mathrm{m}} \\ =\frac{10\left(0.000644\right)}{2.24\mathrm{x}{10}^4\mathrm{M}}=2.9\mathrm{x}{10}^7\mathrm{M} \end{gathered}$$